If dy/dt = ky and k is a nonzero constant, y could be


by lude1
Tags: constant, dy or dt, nonzero
lude1
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#1
May2-10, 12:48 AM
P: 34
1. The problem statement, all variables and given/known data

If dy/dt = ky and k is a nonzero constant, than y could be

a. 2e^kty
b. 2e^kt
c. e^kt + 3
d. kty + 5
e. .5ky^2 + .5

Correct answer is b. 2e^kt

2. Relevant equations



3. The attempt at a solution

I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Squeezebox
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#2
May2-10, 12:54 AM
P: 58
Look at the derivative of e^kt and see how that relates to e^kt.
rock.freak667
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#3
May2-10, 12:56 AM
HW Helper
P: 6,213
The 'correct' thing to do would be

dy/dt = ky

∫ dy/y = ∫ k dt


But since you have a multiple choice, you could quickly differentiate each one if you wanted to.

lude1
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#4
May2-10, 01:07 AM
P: 34

If dy/dt = ky and k is a nonzero constant, y could be


Hm.. this is what I got when I did ∫ dy/y = ∫ k dt

lny = kt
y = e^kt

Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2.
rock.freak667
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#5
May2-10, 01:12 AM
HW Helper
P: 6,213
Quote Quote by lude1 View Post
Hm.. this is what I got when I did ∫ dy/y = ∫ k dt

lny = kt
y = e^kt

Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2.
you'd get

ln y = kt+A such that y=ekt+A=Bekt where B is a constant. So B can be any number, such as '2'.
lude1
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#6
May2-10, 01:14 AM
P: 34
Ah, I forgot about that. Thanks!
HallsofIvy
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#7
May2-10, 08:16 AM
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Quote Quote by lude1 View Post
1. The problem statement, all variables and given/known data

If dy/dt = ky and k is a nonzero constant, than y could be

a. 2e^kty
If this means y=-2e^(kty) then its derivative is dy/dt= 2ke^(kty)(ky+ kt(dy/dt)).
dy/dt(1- 2kt e^(kty)= 2k^y e^(kty). No, that differential equation is not satisfied.

b. 2e^kt
If y= 2e^(kt), then dy/dt= 2(ke^(kt))= k(2e^(kt))= ky and satisfies the equation.
(I am assuming the parentheses combine these as you wanted- that you did not intend 2(e^k)t.)

c. e^kt + 3
If y= e^(kt)+ 3, then y'= ke^(kt) which is not ky= k(e^(kt)+ 3)

d. kty + 5
If y= kty+ 5, then dy/dt= ky+ kt dy/dt so dy/dt(1- kt)= ky and dy/dt= (k/(1-kt) y. That does not satisfy the given equation.

e. .5ky^2 + .5
If y= .5y^2+ .5, then dy/dt= 1.0 y dy/dt so dy/dt(1- y)= 0. Either dy/dt= 0 or y= 1. That does not satisfy the given equation.

Correct answer is b. 2e^kt

2. Relevant equations



3. The attempt at a solution

I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution


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